1)A bus 18.8m long moving at 50km/h overtakes a second bus moving at 45km/h in 22 and 1/2sec.Find the length of the second bus.
2)Kiriari is a market centre 600 km from Kisumu town.A bus starts from Kisumu for Kiriari at 7:00am at an average speed of 80 km/h.At 8:30 am a car started from Kisumu at Kiriaria and moved at an average speed of 120km/h;
a)The distance the bus covered before the car started moving.
b)The relative speed for the two vehicles.
c)The time the car overtook the bus.
d)The distance covered by the car before overtaking the bus.
3)The velocity time graph of an object accelerates from rest to a velocity of V m/s then decelerates to rest in a total time of 54 secs.If the whole journey is 810m.
a)Find the value of V
b)Find the deceleration given the initial acceleration is 1 and 2/3 m/s^2.
4)A car accelerated from rest to a velocity of 10 m/s.It travelled at this velocity for 20 secs and then came to a stop in 5 secs.Find;
a)The initial acceleration
b)The distance travelled.
5)A speed time graph of a journey shows the total distance travelled in 80 secs is 920 m.Calculate;
a)The value of V
b)The acceleration during the first 15 seconds.
c)The distance travelled in the final 40 secs
6)In a triangle P and Q are points on OA and OB respectively such that OP:PA=3:2 and OQ:QB=1:2.AQ and PQ intersect at T.Given that OA=a and OB=b.
a)Express AQ and PQ in terms of a and b.
b)Taking BT=kBP and AT=hAQ where h and k are real numbers;
i)Find two expressions for OT in terms of a and b.
i)Use the expressions in b(i) above to find the values of h and k.
c)Show that B,T and P are collinear
1) Let the length of the second bus be x meters.
The time taken by the first bus to overtake the second bus is 22.5 seconds.
The relative speed between the buses is 50 km/h - 45 km/h = 5 km/h = (5/3.6) m/s.
Using the formula: Relative speed = Distance / Time, we can write:
(5/3.6) = (x + 18.8) / 22.5
Simplifying the equation, we get:
5 = (x + 18.8) * (3.6/22.5)
Solving for x, we find:
x = 6.8 meters
Therefore, the length of the second bus is 6.8 meters.
2) a) The bus started from Kisumu at 7:00am. The car started from Kisumu at 8:30am. Therefore, the time interval between the bus starting and the car starting is 1.5 hours = 1.5 * 60 = 90 minutes = 90 * 60 = 5400 seconds.
The distance covered by the bus before the car started moving is given by:
Distance = Speed * Time = 80 km/h * (5400/3600) h = 120 km.
Therefore, the bus covered a distance of 120 km before the car started moving.
b) The relative speed between the bus and the car is the difference in their speeds.
Relative speed = 120 km/h - 80 km/h = 40 km/h.
c) The car started from Kisumu at 8:30am. The bus started from Kisumu at 7:00am. Therefore, the time interval between the bus starting and the car overtaking the bus is 1.5 hours = 1.5 * 60 = 90 minutes = 90 * 60 = 5400 seconds.
Therefore, the car overtook the bus after 5400 seconds.
d) The distance covered by the car before overtaking the bus can be calculated using the formula:
Distance = Speed * Time = 120 km/h * (5400/3600) h = 180 km.
Therefore, the car covered a distance of 180 km before overtaking the bus.
3) a) The total time taken for the journey is 54 seconds.
Let the initial acceleration be a.
The time taken to accelerate to velocity V is V/a.
The time taken to decelerate from velocity V to rest is V/a.
Therefore, the total time can be written as:
V/a + V/a = 54
Simplifying the equation, we get:
2V/a = 54
Solving for V, we find:
V = 27a
b) The initial acceleration is given as 1 and 2/3 m/s^2.
Therefore, the deceleration can be calculated using the formula:
Deceleration = (V - 0) / (V/a) = V / (V/a) = a
Therefore, the deceleration is 1 and 2/3 m/s^2.
4) The car accelerated from rest to a velocity of 10 m/s.
The time taken to accelerate to velocity 10 m/s is given as 20 seconds.
The initial acceleration can be calculated using the formula:
Acceleration = (V - 0) / Time = (10 - 0) / 20 = 0.5 m/s^2.
The distance travelled at a velocity of 10 m/s for 20 seconds can be calculated using the formula:
Distance = Speed * Time = 10 m/s * 20 s = 200 meters.
5) The total distance travelled in 80 seconds is given as 920 meters.
a) Given that the total distance travelled is equal to the area under the speed-time graph, we can calculate the average speed:
Average speed = Total distance / Total time = 920 m / 80 s = 11.5 m/s.
Therefore, the value of V is 11.5 m/s.
b) The acceleration during the first 15 seconds can be calculated by finding the slope of the speed-time graph during that interval.
The acceleration is the change in speed divided by the change in time:
Acceleration = (10 - 0) / 15 = 2/3 m/s^2.
c) The distance travelled in the final 40 seconds can be calculated by subtracting the distance travelled in the first 40 seconds from the total distance.
The distance travelled in the first 40 seconds can be calculated using the formula:
Distance = (Initial velocity * Time) + (0.5 * Acceleration * Time^2)
= (0 * 40) + (0.5 * (2/3) * 40^2)
= 0 + (0.5 * (2/3) * 1600)
= 0 + (2/3) * 800
= 533.33 meters
Therefore, the distance travelled in the final 40 seconds is 920 - 533.33 = 386.67 meters.
6) a) Let PT = x.
Using the property that the sum of the lengths of two sides of a triangle is greater than the length of the third side, we have:
OA + PT > PA
a + x > (2/3)a
x > (2/3)a - a
x > a/3
Similarly, we have:
OB + TQ > QB
b + x > (1/3)b
x > (1/3)b - b
x > -b/3
Therefore, AQ = a - x > a - a/3 = 2a/3
and PQ = b - x > b - (-b/3) = 4b/3.
b) i) Using the property that triangles OST and AQT are similar, we can write:
OT / AQ = ST / QT
OT / (2a/3) = (a - x) / (b - x)
Simplifying the equation, we get:
OT = (2a/3)(a - x) / (b - x)
ii) Similarly, using the property that triangles OTB and ATP are similar, we can write:
OT / AQ = TB / TP
OT / (2a/3) = (x - b) / x
Simplifying the equation, we get:
OT = (2a/3)(x - b) / x
To find the values of h and k, we can equate the two expressions for OT:
(2a/3)(a - x) / (b - x) = (2a/3)(x - b) / x
Simplifying the equation, we get:
(a - x) / (b - x) = (x - b) / x
Cross-multiplying, we get:
x(a - x) = (x - b)(b - x)
Expanding and simplifying, we get:
ax - x^2 = bx - x^2 - bx + xb
Simplifying further, we get:
ax = 0
Since a is not equal to 0, this implies that x = 0.
Therefore, h = 0 and k = 0.
c) If h = k = 0, then the points B, T, and P are collinear.